1. The problem gives the equation relating pounds of rice $r$ and beans $b$ with total cost $10$: $$2r + 1.60b = 10$$ This equation shows the total cost spent on rice and beans. 2. To express $b$ as the dependent variable (in terms of $r$), rearrange the equation: $$1.60b = 10 - 2r$$ Divide both sides by 1.60: $$b = \frac{10 - 2r}{1.60}$$ Intermediate step with cancellation: $$b = \frac{\cancel{10} - 2r}{\cancel{1.60}}$$ This means $b$ depends on $r$. 3. To express $r$ as the dependent variable (in terms of $b$), rearrange: $$2r = 10 - 1.60b$$ Divide both sides by 2: $$r = \frac{10 - 1.60b}{2}$$ Intermediate step with cancellation: $$r = \frac{\cancel{10} - 1.60b}{\cancel{2}}$$ This means $r$ depends on $b$. 4. Now solve each of the three equations given: (a) Solve $$2x + 4(3 - 2x) = \frac{3(2x+2)}{6} + 4$$ Step 1: Expand and simplify both sides: $$2x + 12 - 8x = \frac{6x + 6}{6} + 4$$ $$2x + 12 - 8x = x + 1 + 4$$ $$-6x + 12 = x + 5$$ Step 2: Bring all $x$ terms to one side: $$-6x - x = 5 - 12$$ $$-7x = -7$$ Step 3: Divide both sides by $-7$: $$x = \frac{-7}{-7} = 1$$ Check: Substitute $x=1$ back into original equation: Left: $2(1) + 4(3 - 2(1)) = 2 + 4(1) = 2 + 4 = 6$ Right: $\frac{3(2(1)+2)}{6} + 4 = \frac{3(4)}{6} + 4 = 2 + 4 = 6$ Both sides equal 6, so $x=1$ is correct. (b) Solve $$4z + 5 = -3z - 8$$ Step 1: Bring $z$ terms to one side and constants to the other: $$4z + 3z = -8 - 5$$ $$7z = -13$$ Step 2: Divide both sides by 7: $$z = \frac{-13}{7}$$ Check: Substitute $z = -\frac{13}{7}$: Left: $4(-\frac{13}{7}) + 5 = -\frac{52}{7} + 5 = -\frac{52}{7} + \frac{35}{7} = -\frac{17}{7}$ Right: $-3(-\frac{13}{7}) - 8 = \frac{39}{7} - 8 = \frac{39}{7} - \frac{56}{7} = -\frac{17}{7}$ Both sides equal $-\frac{17}{7}$, so solution is correct. (c) Solve $$\frac{1}{2} - \frac{1}{8}q = \frac{q - 1}{4}$$ Step 1: Multiply entire equation by 8 to clear denominators: $$8 \times \left( \frac{1}{2} - \frac{1}{8}q \right) = 8 \times \frac{q - 1}{4}$$ $$4 - q = 2(q - 1)$$ Step 2: Expand right side: $$4 - q = 2q - 2$$ Step 3: Bring $q$ terms to one side and constants to the other: $$4 + 2 = 2q + q$$ $$6 = 3q$$ Step 4: Divide both sides by 3: $$q = 2$$ Check: Substitute $q=2$: Left: $\frac{1}{2} - \frac{1}{8} \times 2 = \frac{1}{2} - \frac{2}{8} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$ Right: $\frac{2 - 1}{4} = \frac{1}{4}$ Both sides equal $\frac{1}{4}$, so solution is correct.